Arkansas Tech University
MATH 4033: Elementary Modern Algebra
Dr. Marcel B. Finan
22
Quotient Groups
Let’s look closely on the construction of the group (ZZn , ⊕). We know that the
elements in ZZn are equivalence classes of the equivalence relation defined on ZZ
by
a ∼ b if and only if n|(a − b).
Also, an element of ZZn is a right coset. Indeed, if 0 ≤ k < n then
< n > +k = [k] = {nq + k : q ∈ ZZ}.
The operation ⊕ is defined by
[a] ⊕ [b] = [a + b]
or using right cosets
(< n > +a) ⊕ (< n > +b) =< n > +(a + b).
Note that the cyclic group < n > is a normal subgroup of ZZ since ZZ is Abelian
(See Example 21.4). We are going to use the above ideas to construct new groups
where G is a group replacing ZZ and N is a normal subgroup of G playing the
role of < n > . More precisely, we have the following theorem.
Theorem 22.1
Let N be a normal subgroup of a group G and let G/N be the set of all right
cosets of N in G. Define the operation on G/N × G/N by
(N a)(N b) = N (ab).
Then (G/N, ·) is a group, called the quotient group of G by N.
Proof.
· is a well-defined operation
We must show that if (N a1 , N b1 ) = (N a2 , N b2 ) then N (a1 b1 ) = N (a2 b2 ). To
see this, since (N a1 , N b1 ) = (N a2 , N b2 ) then N a1 = N a2 and N b1 = N b2 .
Since a1 = ea1 ∈ N a1 then a1 ∈ N a2 so that a1 = na2 for some n ∈ N. Similarly, b1 = n0 b2 for some n0 ∈ N. Therefore, a1 b1 = na2 n0 b2 . Since N / G
then a2 n0 a−1
∈ N, say a2 n0 a−1
= n00 ∈ N. Hence, a2 n0 = n00 a2 so that
2
2
00
00
a1 b1 = nn a2 b2 . But nn ∈ N so that a1 b1 ∈ N (a2 b2 ). Since N (a2 b2 ) is an
equivalence class and a1 b1 ∈ N (a2 b2 ) then N (a1 b1 ) = N (a2 b2 ) (Theorem 9.2).
· is associative
Let a, b, c ∈ G. Then
N a(N bN c) = N a(N bc) = N (a(bc)) = N ((ab)c) = N (ab)N c = (N aN b)(N c)
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where we used the fact that multiplication in G is associative.
N e = N is the identity element
If a ∈ G then (N a)(N e) = N (ae) = N a and (N e)(N a) = N (ea) = N a.
Every element of G/N is invertible
If a ∈ G then (N a)(N a−1 ) = N (aa−1 ) = N e and (N a−1 )(N a) = N (a−1 a) =
N e so that (N a)−1 = N a−1 .
Remark 22.1
Note that for a finite group G, the number of elements of G/N is just the index
of N in G, i.e. [G : N ]. That is, |G/N | = [G : N ]. By Lagrange’s theorem,
|G|
|G|
|G| = [G : N ]|N | so that [G : N ] = |N
| . Hence, |G/N | = |N | .
Remark 22.2
If N is not a normal subgroup of G, then the operation defined in the previous theorem will not be well-defined. To see this, consider the subgroup
N =< (12) >= {(1), (12)} of S3 . Since (123)(12)(123)−1 = (23) 6∈ N then
N is not a normal subgroup of S3 . However, N (123) = N (23) = {(123), (23)}
and N (132) = N (13) = {(13), (132)}. But N (123)(132) 6= N (23)(13) since
N (123)(132) = N and N (23)(13) = N (123).
Example 22.1
Let G = S3 and N =< (123) >= {(1), (123), (132)}. One can verify easily that
N / G, G/N = {N, N (12)}.
Example 22.2
Quotient groups can be used to show that A4 has no subgroup of order 6 and
thus showing that the converse of Lagrange’s Theorem is false in general. To
see this, assume that N is a subgroup of A4 of order 6. Then [G : N ] = 2 and
therefore N / A4 (See Exercise 21.7). Hence, A4 /N makes sense. Moreover, for
each a ∈ N, (N a)−1 = N a so that N a2 = N and hence a2 inN. One can show
that (123)2 = (132), (132)2 = (123), (124)2 = (142), (142)2 = (124), (134)2 =
(143), (143)2 = (134), (234)2 = (243), and (243)2 = (234). This yields more
than six different elements of A4 in N. That is, |N | > 6, a contradiction. Thus,
A4 has no subgroup of order 6.
By Theorem 21.1(ii), the kernel of a homomorphism is a normal subgroup.
Conversely, quotient groups enable us to show that every normal subgroup is
the kernel of some homomorphism.
Theorem 22.2
Let G be a group and let N / G. The mapping η : G −→ G/N defined by
η(a) = N a for all a ∈ G is a homomorphism such that Ker η = N. We call η
the natural homomorphism from G onto G/N.
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Proof.
First we show that η is well-defined. Indeed, if a = b then N a = N b, i.e.
η(a) = η(b). Since η(ab) = N (ab) = N aN b = η(a)η(b) then η is a homomorphism. η is onto since any member of G/N is of the form N a for some a ∈ G.
That is, η(a) = N a. It remains to show that Ker η = N. The proof is by double
inclusions. If a ∈ Ker η then η(a) = N e. Thus, N a = N e so that a = ne = n
for some n ∈ N. Hence, a ∈ N and Ker η ⊆ N. Conversely, for all n ∈ N we
have η(n) = N n = N = N e so that N ⊆ Ker η.
The operation of multiplication on the quotient group was defined in terms
of right cosets. The following theorem shows that when working with cosets of
a normal subgroup N , it is immaterial whether we use N a or aN.
Example 22.3
If G = ZZ and N =< n > then G/N = ZZn and η : ZZ −→ ZZn is given by
η(a) =< n > +a = [a] and Ker η =< n > .
Theorem 22.3
Let N be a normal subgroup of a group G. Then aN = N a for all a ∈ G.
Proof.
Let x ∈ aN. Then x = an for some n ∈ N. Since N is normal then ana−1 ∈ N.
Thus, ana−1 = n0 ∈ N and therefore an = n0 a. That is, x ∈ N a. Hence,
aN ⊆ N a. A similar argument shows that N a ⊆ aN.
We close this section with some properties of quotient groups. More properties of quotient groups are discussed in the exercises.
Lemma 22.1
If G is Abelian and N / G then G/N is also Abelian.
Proof.
We have to show that for any N a, N b ∈ G/N we have (N a)(N b) = (N b)(N a).
Now, since G is Abelian then ab = ba. Thus,
(N a)(N b) = N ab = N ba = (N b)(N a).
That is, G/N is Abelian.
Lemma 22.2
If G is a group such that G/Z(G) is cyclic then G is Abelian, where
Z(G) = {g ∈ G : xg = gx ∀x ∈ G}.
Proof.
First we show that Z(G) / G. Indeed, if g ∈ G and x ∈ Z(G) then gxg −1 =
gg −1 x = x ∈ Z(G). Since G/Z(G) is a cyclic group then G/Z(G) =< Z(G)g >
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for some g ∈ G. If a, b ∈ G then Z(G)a and Z(G)b belong to G/Z(G). Thus,
Z(G)a = Z(G)g n and Z(G)b = Z(G)g m for some integers n and m. Hence, a =
xg n and b = yg m for some x, y ∈ Z(G). This implies that ab = (xg n )(yg m ) =
xyg n+m = yxg m g n = yg m xg n = ba. Hence, G is Abelian.
Lemma 22.3
Let G be a group of order p2 where p is a prime number. Assuming that the
center of G is nontrivial, then G is Abelian.
Proof.
Let G be a group such that |G| = p2 . Since e ∈ C(G) then C(G) 6= ∅. By
Lagrange’s theorem either |C(G)| = p2 or |C(G)| = p. If |C(G)| = p2 then
C(G) = G and so G is Abelian. If |C(G)| = p then |G/C(G)| = p and so
G/C(G) is cyclic (See Corollary 3). By the previous lemma, G is Abelian.
Review Problems
Exercise 22.1
Find the order of each of the following quotient groups.
(i) ZZ8 / < [4] > .
(ii) < 2 > / < 8 > .
Exercise 22.2
Let G = A4 and H = {(1), (12)(34), (13)(24), (14)(23)} / G. Write out the distinct elements of G/H and make a Cayley table for G/H.
Exercise 22.3
Construct the Cayley table for ZZ12 / < [4] > .
Exercise 22.4
Let G be a cyclic group. Prove that for every subgroup H of G, G/H is cyclic.
Exercise 22.5
Assume N / G.
(a) Prove that if [G : N ] is prime, then G/N is cyclic.
(b) Prove or disprove the converse of the statement in part (a).
Exercise 22.6
Determine the order of ZZ12 × ZZ4 / < [3], [2] > .
Exercise 22.7
Prove that if N / G and a ∈ G then o(N a)|o(a).
Exercise 22.8
Prove that G/N is Abelian if and only if aba−1 b−1 ∈ N for all a, b ∈ G.
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Exercise 22.9
Let N be a normal subgroup of G and a ∈ G. Prove that the order of N a in
G/N is the smallest positive integer n such that an ∈ N.
Exercise 22.10
Let θ : G → H be a homomorphism and K = Ker θ. Show that φ : G/K → θ(G)
defined by φ(Ka) = θ(a) is an isomorphism. Also, show that φ ◦ η = θ, where η
is the canonical homomorphism.
Exercise 22.11
Let H and K be subgroups of a group G such that K / G. Prove that H/H ∩ K
and HK/K make sense, where HK = {hk : h ∈ H and k ∈ K}.
Exercise 22.12
Let H and K be as in the previous exercise. Show that φ : H → HK/K defined
by φ(h) = Kh an epimorphism with Ker φ = H ∩ K.
Exercise 22.13
Assume that H, K4G and K4H. Prove that H/K / G/K.
Exercise 22.14
Let H, K, and G as in the previous exercise. Show that φ : G/K → G/H,
defined by φ(Kg) = Hg is an epimorphism with Ker φ = H/K.
Exercise 22.15
If N is a subgroup of G such that the product of two right cosets of N in G is
again a right coset of N in G, prove that N / G.
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